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λ = h p = h c K ( K + 2 E 0 ) .

Depending on the problem at hand, in this equation we can use the following values for hc : h c = ( 6.626 × 10 −34 J · s ) ( 2.998 × 10 8 m/s ) = 1.986 × 10 −25 J · m = 1.241 eV · μ m


  1. For the basketball, the kinetic energy is
    K = m 0 u 2 / 2 = ( 0.65 kg ) ( 10 m/s ) 2 / 2 = 32.5 J

    and the rest mass energy is
    E 0 = m 0 c 2 = ( 0.65 kg ) ( 2.998 × 10 8 m/s ) 2 = 5.84 × 10 16 J.

    We see that K / ( K + E 0 ) 1 and use p = m 0 u = ( 0.65 kg ) ( 10 m/s ) = 6.5 J · s/m :
    λ = h p = 6.626 × 10 −34 J · s 6.5 J · s/m = 1.02 × 10 −34 m .
  2. For the nonrelativistic electron,
    E 0 = m 0 c 2 = ( 9.109 × 10 −31 kg ) ( 2.998 × 10 8 m/s ) 2 = 511 keV

    and when K = 1.0 eV , we have K / ( K + E 0 ) = ( 1 / 512 ) × 10 −3 1 , so we can use the nonrelativistic formula. However, it is simpler here to use [link] :
    λ = h p = h c K ( K + 2 E 0 ) = 1.241 eV · μ m ( 1.0 eV ) [ 1.0 eV+ 2 ( 511 keV ) ] = 1.23 nm .

    If we use nonrelativistic momentum, we obtain the same result because 1 eV is much smaller than the rest mass of the electron.
  3. For a fast electron with K = 108 keV, relativistic effects cannot be neglected because its total energy is E = K + E 0 = 108 keV + 511 keV = 619 keV and K / E = 108 / 619 is not negligible:
    λ = h p = h c K ( K + 2 E 0 ) = 1.241 eV · μm 108 keV [ 108 keV + 2 ( 511 keV ) ] = 3.55 pm .


We see from these estimates that De Broglie’s wavelengths of macroscopic objects such as a ball are immeasurably small. Therefore, even if they exist, they are not detectable and do not affect the motion of macroscopic objects.

Check Your Understanding What is de Broglie’s wavelength of a nonrelativistic proton with a kinetic energy of 1.0 eV?

1.7 pm

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Using the concept of the electron matter wave, de Broglie provided a rationale for the quantization of the electron’s angular momentum in the hydrogen atom, which was postulated in Bohr’s quantum theory. The physical explanation for the first Bohr quantization condition comes naturally when we assume that an electron in a hydrogen atom behaves not like a particle but like a wave. To see it clearly, imagine a stretched guitar string that is clamped at both ends and vibrates in one of its normal modes. If the length of the string is l ( [link] ), the wavelengths of these vibrations cannot be arbitrary but must be such that an integer k number of half-wavelengths λ / 2 fit exactly on the distance l between the ends. This is the condition l = k λ / 2 for a standing wave on a string. Now suppose that instead of having the string clamped at the walls, we bend its length into a circle and fasten its ends to each other. This produces a circular string that vibrates in normal modes, satisfying the same standing-wave condition, but the number of half-wavelengths must now be an even number k , k = 2 n , and the length l is now connected to the radius r n of the circle. This means that the radii are not arbitrary but must satisfy the following standing-wave condition:

2 π r n = 2 n λ 2 .

If an electron in the n th Bohr orbit moves as a wave, by [link] its wavelength must be equal to λ = 2 π r n / n . Assuming that [link] is valid, the electron wave of this wavelength corresponds to the electron’s linear momentum, p = h / λ = n h / ( 2 π r n ) = n / r n . In a circular orbit, therefore, the electron’s angular momentum must be

L n = r n p = r n n r n = n .

This equation is the first of Bohr’s quantization conditions, given by [link] . Providing a physical explanation for Bohr’s quantization condition is a convincing theoretical argument for the existence of matter waves.

Questions & Answers

A round diaphragm S with diameter of d = 0.05 is used as light source in Michelson interferometer shown on the picture. The diaphragm is illuminated by parallel beam of monochromatic light with wavelength of λ = 0.6 μm. The distances are A B = 30, A C = 10 . The interference picture is in the form of concentric circles and is observed on the screen placed in the focal plane of the lens. Estimate the number of interference rings m observed near the main diffractive maximum.
Jyoti Reply
A Pb wire wound in a tight solenoid of diameter of 4.0 mm is cooled to a temperature of 5.0 K. The wire is connected in series with a 50-Ωresistor and a variable source of emf. As the emf is increased, what value does it have when the superconductivity of the wire is destroyed?
Rupal Reply
how does colour appear in thin films
Nwjwr Reply
in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
Kimani Reply
derivation of lateral shieft
James Reply
total binding energy of ionic crystal at equilibrium is
All Reply
How does, ray of light coming form focus, behaves in concave mirror after refraction?
Bishesh Reply
Refraction does not occur in concave mirror. If refraction occurs then I don't know about this.
What is motion
Izevbogie Reply
Anything which changes itself with respect to time or surrounding
and what's time? is time everywhere same
how can u say that
do u know about black hole
Not so more
Radioactive substance
These substance create harmful radiation like alpha particle radiation, beta particle radiation, gamma particle radiation
But ask anything changes itself with respect to time or surrounding A Not any harmful radiation
explain cavendish experiment to determine the value of gravitational concept.
Celine Reply
 Cavendish Experiment to Measure Gravitational Constant. ... This experiment used a torsion balance device to attract lead balls together, measuring the torque on a wire and equating it to the gravitational force between the balls. Then by a complex derivation, the value of G was determined.
For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?
Evan Reply
as a free falling object increases speed what is happening to the acceleration
Success Reply
of course g is constant
acceleration also inc
which paper will be subjective and which one objective
normal distributiin of errors report
normal distribution of errors
acceleration also increases
there are two correct answers depending on whether air resistance is considered. none of those answers have acceleration increasing.
Acceleration is the change in velocity over time, hence it's the derivative of the velocity with respect to time. So this case would depend on the velocity. More specifically the change in velocity in the system.
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
Rafi Reply
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
Mohammed Reply
give any fix value to wave length
40 cm into change mm
Arhaan Reply
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
this msg is out of mistake. sorry friends​.
what is physics?
sisay Reply
Practice Key Terms 5

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